jmc

Let $J$ be the intersection of $\overline{BE}$ and $\overline{GC}$.

Observe that $BD=\sqrt{B{E}^2-D{E}^2}=\sqrt{5^2-3^2}=4$ units. By the similarity of triangles $BCJ$ and $BDE$, we have $$\frac{CJ}{BC}=\frac{DE}{BD},$$ which after substituting becomes $$\frac{CJ}{4-3}=\frac{3}{4}.$$ We solve to find $CJ=\frac{3}{4}$, which implies that $GJ=3-\frac{3}{4}=\frac{9}{4}$. Applying the Pythagorean theorem to triangle $GJE$, we find $EJ=\sqrt{3^2+{\left(\frac{9}{4}\right)}^2}=\frac{15}{4}$. Define $K$ to be the foot of the perpendicular from $G$ to side $EJ$. By the similarity of triangles $GKJ$ and $EGJ$, we have $$\frac{GK}{GJ}=\frac{EG}{EJ}⟹\frac{GK}{\frac{9}{4}}=\frac{3}{\frac{15}{4}},$$ which we solve to find $GK=\frac{9}{5}$. Since $GKBH$ is a rectangle, $BH=GK=\frac{9}{5}=\boxed{1\frac{4}{5}}$ units.